Assume that \(\left( \Phi _{nf}^{\ast }(p),k+1\right) >1\) . Then \(X_{x}\)  contains a Singer cyclic subgroup of \(X\)  by
Lemma REF . Let \(Y\)  be a maximal subgroup of \(X\)  containing \(X_{x}\) . It follows from the Main Theorem of [1]} that, either \(Y\in \mathcal {C}_{3}(X)\)  or, since \(\Phi _{nf}^{\ast }(p)>1\)  and \(\Phi _{nf}^{\ast }(p)\mid \left|Y\right|\) , one of the following holds:

\(X\cong PSU_{n}(q)\) , where \((n,q)=(3,3),(3,5)\)  or \((5,2)\) .

\(X\cong PSp_{n}(q)\) , where either \(q\)  is even and \(\Omega _{n}^{-}(q)\leqslant Y\) , or \((n,q)=(8,2)\) .

\(X\cong P\Omega _{6}^{-}(3)\) .

The case \(X\cong PSU_{3}(3)\)  is ruled out in Lemma 3.6 of [2]}, as \(PSU_{3}(3)\cong G_{2}(2)^{\prime }\) . The group \(PSU_{3}(5)\) 
is ruled out by [3]}, as it does not have a transitive permutation representation of
square degree. The same conclusion holds for \(PSp_{8}(2)\)  by [4]},
Tables 8.48–8.49. If \(X\)  is isomorphic either to \(PSU_{5}(2)\)  or to \(P\Omega _{6}^{-}(3)\) , then \(X_{x}\cong PSL_{2}(11)\)  and \(k=2^{4}\cdot 3^{2}\) , or \(X_{x}\cong A_{7}\)  and \(k=2^{4}\cdot 3^{2}\)  , respectively, by [3]}. However, these cases are ruled out since \(k+1\)  does not divide the order
of \(\mathrm {Aut(}X\mathrm {)}\)  and hence that of \(G\) .
Suppose that \(X\cong PSp_{n}(q)\) , with \(q\)  even, and \(\Omega _{n}^{-}(q)\leqslant Y\) . Then \(Y\cong SO_{n}^{-}(q)\)  by [6]}, Proposition
4.8.6. If \(X_{x} \ne Y\) , then \(X_{x}\cong \Omega _{n}^{-}(q)\)  again by Main Theorem of [1]} but applied to \(Y\) . Thus, \(X_{x}\)  is isomorphic either to \(\Omega _{n}^{-}(q)\)  or to \(SO _{n}^{-}(q)\) . If \(\rho \)  is the index of \(X_{x}\)  in \(Y\) , then \(\rho =1,2\)  and hence \(k^{2}=\frac{q^{n(n+2)/4}}{\rho }(q^{n/2}-1)\) , which has no solutions by [8]}, A3.1, since \(n\geqslant 4\)  by Remark REF . Thus, \(Y\in \mathcal {C}_{3}(X)\) .
The group \(X\)  contains a unique conjugacy class of subgroups isomorphic to \(Y \) , except for \(X\cong P\Omega _{n}^{-}(q)\)  and \(O_{n/2}(q^{2}) \)  with \(qn/2\)  odd by [6]}, Section 4.3, since \(\Phi _{nf}^{\ast }(p)\mid \left|X_{x}\right|\)  and since \(n \geqslant 6\)  (see Proposition 4.3.20.(I) for the exception). Then \(X_{x}=Y\)  by Lemma REF (1) when \(Y\)  is none of the
exceptions. In the exceptional cases, \(X_{x}\)  is a type 1 novelty with
respect to \(Y\)  by Lemma REF (2). However, this is impossible by [6]}, Tables 3.5.H–I, for \(n\geqslant 13\)  and [4]} for \(3\leqslant n\leqslant 12\) .
Thus \(X_{x}=Y\)  and hence \(X_{x}\)  is a maximal \(\mathcal {C}_{3}\) -subgroup of \(X\) .
Suppose that \(X_{x}\)  is not a large subgroup of \(X\) . Then \(\left|X_{x}\right|_{p}<\left|\mathrm {Out(}X\mathrm {)}\right|\)  by
Lemma REF . Hence, either \(X\cong PSL_{n}(q)\)  and \(X_{x}\)  is of type \(GL_{1}(q^{n})\) , or \(X\cong PSU_{n}(q^{1/2})\) , with \(n\)  odd, and \(X_{x}\)  is
of type \(GU_{1}(q^{n/2})\) , or \(X\cong P\Omega _{6}^{-}(q)\)  by [12]},
Lemma 3.13, since \(\Phi _{nf}^{\ast }(p)\mid \left|X_{x}\right|\) .
Assume that \(X\cong PSL_{n}(q)\)  and \(X_{x}\)  is of type \(GL_{1}(q^{n})\) . Then
\(q^{n^{2}-2}<\left|X\right|\leqslant \left|G\right|\leqslant \left( (q^{n}-1)nf\right) ^{3}\leqslant q^{3n+3}n^{3}\) 
since \(G_{x}\)  is a large subgroup of \(G\)  normalizing a Singer cyclic subgroup of \(X\)  and by [13]}, Corollary 4.3.(i). Hence \(n=3,4\) . If \(n=4\) , then
\(k^{2}=\frac{q^{6}(q-1)^{2}(q^{2}-1)(q^{2}+q+1)}{4}\) 
by [6]}, Proposition 4.3.6.(II). On the other hand, \(\frac{k+1}{(k+1,f)}\mid 4 \frac{q^{4}-1}{(q-1)(4,q-1)} \)  by
Lemma REF , and hence \(k+1\leqslant 4f\frac{q^{4}-1}{(q-1)(4,q-1)}\) , which, compared to (REF ), yields \(q \leqslant 5\) . However, none of these values fulfills (REF ). Thus \(n=3\)  and hence \(\left|X_{x}\right|=3 \frac{q^{2}+q+1}{(3,q-1)}\)  again by [6]}, Proposition 4.3.6.(II). Therefore,
\(k^{2}=\frac{q^{3}(q-1)^{2}(q+1)}{3 }.\) 
If \(q\equiv 1 \pmod {3}\) , then both \(q\)  and \(q+1\)  are squares, but this contradicts
[8]}, A5.1. If \(q\equiv 0 \pmod {3}\) , then both \(q^{3}/3\)  and \(q+1\)  are squares
and hence \(q=3\)  again by [8]}, A5.1. So \(\left|X_{x}\right|_{p}=3\) , whereas \(\left|\mathrm {Out(}X\mathrm {)}\right|=2\) . Therefore, \(q\equiv 2 \pmod {3}\)  and hence both \(q\)  and \(\frac{q+1}{3}\)  are squares, but this is impossible by [8]}, P5.5.
Assume \(X\cong PSU_{n}(q^{1/2})\) , with \(n\)  odd, and \(X_{x}\)  is
of type \(GU_{1}(q^{n/2})\) . Then
\((q^{1/2}-1)q^{\frac{n^{2}-3}{2}}<\left|X\right|\leqslant \left|G\right|\leqslant \left((q^{n/2}+1)nf\right) ^{3}\leqslant (q^{n/2}+1)^{3}q^{3/2}n^{3}\) 
since \(G_{x}\)  is a large subgroup of \(G\)  normalizing a Singer cyclic subgroup of \(X\) , since \(f\)  is even and hence \(q^{1/2}\geqslant f\) , and by [13]}, Corollary 4.3.(ii). Thus either \(n=3\) , or \(n=5\)  and \(q^{1/2}=2\) , since \(n\)  is odd. We have seen above that \(PSU_{5}(2)\)  cannot occur, hence the latter is ruled out. Thus \(n=3\)  and hence \(\left|X_{x}\right|=3 \frac{q-q^{1/2}+1}{(3,q^{1/2}+1)}\)  and
\(k^{2}=\frac{q^{3/2}(q^{1/2}-1)(q^{1/2}+1)^{2}}{3 }\text{.}\) 
by [6]}, Proposition 4.3.6.(II). Then \(q^{1/2}\equiv 0,1 \pmod {3} \)  by [8]}, A3.1. Since \(\frac{k+1}{(k+1,f)}\mid 3\left( q-q^{1/2}+1\right) \)  by
Lemma REF , it follows that \(k+1\leqslant 3f\left( q-q^{1/2}+1\right)\) , which, compared to (REF ), yields \(q^{1/2}=3,4\) . However, none of these fulfills (REF ).
Finally, assume that \(X\cong P\Omega _{6}^{-}(q)\) . Then \(\left|\mathrm {Out(}X\mathrm {)}\right|=(4,q^{3}+1)2f\)  and \(X_{x}\) 
is a subgroup of \(\frac{1}{(2,q-1)}GU_{3}(q)\)  by [6]}, Propositions
4.3.18.(II) and 4.3.20.(II), as \(\Phi _{6f}^{\ast }(p)\mid \left|X_{x}\right|\) . Since \(q\mid \left|X_{x}\right|_{p}\)  by Corollary REF (v), and since \(\left|X_{x}\right|_{p}<(4,q^{3}+1)2f\) , it follows that \(q<8f\)  and hence \(q=2,3,5\)  or 7. Actually, \(q\ne 2,3\)  as \(P\Omega _{6}^{-}(2)\)  is ruled out in Lemma REF , since \(\Phi _{6}^{\ast }(2)=1\) , and \(P\Omega _{6}^{-}(3)\)  is ruled out above in
this proof. Since \(\left[ X:X_{x}\right] =k^{2}\) , it is not difficult to obtain that \(2\cdot 7\cdot 13\)  or 43
divides \(\left|X_{x}\right|\)  according to whether \(q=5\)  or 7
respectively. However, no cases arise by [4]}, Tables 8.10–8.11,
since \(X_{x}\)  is a maximal \(\mathcal {C}_{3}\) -subgroup of \(X\) .
